Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that for $n\ge 2$, $n\in \mathbb{N}$, $a_{i}, b_{i}\ge 0$

$$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n}a^2_{i}\right)+\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n}b_{i}\right)^2\ge\displaystyle\prod_{i=1}^{n}(a^2_{i}+b^2_{i})^{\frac{1}{n}}$$

When $n=2$, I can easily show that

$$a^2_{1}+a^2_{2}+\dfrac{1}{4}(b_{1}+b_{2})^2\ge\sqrt{(a^2_{1}+b^2_{1})(a^2_{2}+b^2_{2})}$$

I want to use Induction to prove this. Any suggestions on how to solve it?

share|improve this question
    
I don't think you should use induction on this. –  Euclidean Mar 25 '13 at 7:20

1 Answer 1

up vote 2 down vote accepted

Fix $a_2,\cdots,a_n, b_2,\cdots, b_n$. Consider $b_1' = t$, $a_1'^2 = a_1^2 + b_1^2 - t^2$, for $t \in [0,\sqrt{a_1^2+b_1^2}]$. With this pair $(a_1',b_1')$ replacing $(a_1,b_1)$, one sees that right hand side is unchanged, and left hand side is a concave quadratic function in $t$. This means that LHS attains its minimum when $t$ hits the boundary, i.e. $(a_1,b_1)$ being replaced by $(\sqrt{a_1^2+b_1^2}, 0)$, or $(0, \sqrt{a_1^2+b_1^2})$.

This reduces the problem to the following one. Let $x_i^2 = a_i^2 + b_i^2$, $x_i \ge 0$. For each index $i$, $(a_i,b_i) = (x_i,0)$ or $(0,x_i)$. Rearrange the indices so that the first $k$ indices satisfy $(a_i,b_i) = (x_i,0)$, and the last $(n-k)$ satisfies $(a_i,b_i) = (0,x_i)$. We then need to show

$$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{k}x_i^2\right)+\left(\dfrac{1}{n}\displaystyle\sum_{i=k+1}^{n}x_i\right)^2\ge\displaystyle\prod_{i=1}^{n}x_i^{\frac{2}{n}}$$

for any $x_i \ge 0$.

Apply AM-GM for the two terms on the left separately, we get LHS is at least

$$\frac{k}{n} \left(\dfrac{n}{n-1}\right)^{n-1} (x_1\cdots x_k)^{1/k} + \frac{(n-k)^2}{n^2} (x_{k+1}\cdots x_n)^{1/(n-k)}$$

The inequality is clearly true when $k = 0$ or $n$. Otherwise apply AMGM again, with suitable weights, we get that this is at least $$\left(\dfrac{n}{n-1}\right)^{k(n-1)/n}\left(\frac{n-k}{n}\right)^{(n-k)/n}(x_1\cdots x_n)^{2/n}$$ So it suffices to show that $$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\frac{n-k}{n}\right)^{(n-k)/k} \ge 1$$ Equivalently, $$\left(\dfrac{n}{n-1}\right)^{n-1} \ge \left(\frac{n}{n-k}\right)^{(n-k)/k} = \left(\frac{1}{1 - k/n}\right)^{n/k - 1}$$ Hence consider the function $f(u) = (1-u)^{u-1}$ for $u \in (0,1)$. It suffices to show that $f(u)$ is increasing in this range, which is true from the plot at wolframalpha.

share|improve this answer
    
Oh,very nice, your methods Now I can't find any mistake, Thank you very much,and other have any idea? –  math110 Mar 25 '13 at 10:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.